Chemistry Homework Answer Key #6
Chapter 10 - Acid-Base Equilibria
71.
Recall that in this problem, I adapted Ka1 to be 1.00 ´ 10-5 and Ka2 to be 1.00 ´ 10-9; the acid is simply a generic H2A species. The volumes and concentrations remain the same as in the problem.
In the titration of H2A with NaOH, the reaction preceding the first endpoint is the neutralization of hydroxide with the first proton of H2A, or,
It is important to realize that the product of the above reaction is amphoteric - it is the conjugate base of H2A, but it can also act as an acid, albeit a much weaker one than H2A. Because it can act as an acid, it will react with additional hydroxide as shown below.
The second endpoint of the titration is seen when this second reaction is completed. Because the first reaction produces an equivalent number of moles of HA- as H2A that was initially present, it will take just as much NaOH solution to go from the first to the second endpoint as it will reach the first endpoint. That is, the second endpoint will occur at twice the volume of NaOH added as the first.The first endpoint will occur at:
The second endpoint will therefore be seen at twice this value, or 100.0 mL.
a) The pH before any NaOH is added is gotten by making the simplifying assumption that all of the hydronium present at equilibrium is due to the first ionization process. That is, we can treat the diprotic acid as a monoprotic acid with a Ka equal to Ka1 of the diprotic acid. This then becomes a simple weak acid problem. The reaction is:
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x = 1.00 ´ 10-3
Since x = [H3O+], pH = -log (1.00 ´ 10-3) = 3.00
b) after 5.00 mL of NaOH are added:Before the first endpoint the diprotic acid is being converted to its conjugate base, the amphoteric HA-, by the reaction,
OH- (aq) + H2A (aq) ® H2O + HA- (aq)
Since the solution contains both the acid and the conjugate base, we can use the Henderson-Hasselbach equation. To keep track of the relative levels of the acid and conjugate base, it helpful to set up a table of moles; the number of moles of hydroxide is obtained by multiplying the volume by the concentration.
moles H2A OH- HA- initial 5.000 x 10-3 0.500 x 10-3 0 change -0.500 x 10-3 -0.500 x 10-3 0.500 x 10-3 equilibrium 4.500 x 10-3 0 0.500 x 10-3 Using these values in the Henderson Hasselbach equation yields,
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c) after 25.00 mL of NaOH are added:
This is still prior to the first endpoint and the problem is solved just as in part b, above.
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Using these values in the Henderson Hasselbach equation yields,
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d) after 50.00 mL of NaOH are added:
This is exactly at the first endpoint. Thus, all of the H2A has been converted to HA-, but, none of the HA- has been converted to A2-. Since HA- is amphoteric, it can react as both an acid and a base according to the following reactions.
HA- + H2O « H2A + OH-
As we discussed in class, the pH of an amphoteric compound can be estimated using the relationship:
Note that the pH is 7 only because of the values of Ka1 and Ka2; it a general rule that the pH of amphoteric compounds is the same as pure water.
e) after 75.00 mL of NaOH are added:Between the first and second endpoints, the amphoteric HA- (the product of the reaction occurring prior to the first endpoint) is acting as an acid and is being converted to its conjugate base, A2-, by the reaction,
OH- (aq) + HA- (aq) ® H2O + A2- (aq)
Since the solution contains both the acid and the conjugate base, we can use the Henderson-Hasselbach equation. In this case you need to use pKa2 instead of pKa1. In the table of moles present, only concern yourself with the moles of hydroxide added after the first endpoint; this is because all of the hydroxide added before the first endpoint was used to generate the HA-.
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Using these values in the Henderson Hasselbach equation yields,
f) after 99.9 mL of NaOH are added:
This is still between the first and second enpoints, so it should be possible to solve the problem exactly as in part e above. One would think....
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Using these values in the Henderson Hasselbach equation yields,
As part g will indicate, this is an overestimate of the pH. The problem is that when the ratio of A-/HA becomes too great or too small, the assumptions inherent in the table above become problematic. For example, if the amount of OH- added was 4.9999999 ´ 10-3, the above calculation would result in a A-/HA ratio of 1 ´ 1010, giving a pH of 19.00. This is clearly wrong. As it turns out, at the endpoint, the concentration of HA- that arises from the base equilibrium reaction of A2- is 5.7 ´ 10-4. This is roughly 10 times the value that the above table indicates. So, a word of warning, do not use the type of table above, i.e., those that assume complete stoichiometric conversion of reactants and neglect equilibrium considerations, when too close to either the beginning or end of a titration curve. As it turns out, the actual pH would be 10.73, so the answer above represents a substantial error.
g) after 100.0 mL of NaOH are added:
This is at the second endpoint. Therefore, all of the HA- has been converted to A2- and no excess OH- has been added. Just as the fully acidic form, H2A, was treated as a monoprotic acid with Ka = Ka1, the fully basic form can be treated as a monoprotic base with Kb = Kb1.
This then becomes a simple weak base problem. The reaction is:
A2- + H2O « HA- + OH-
concentration A2- HA- OH- initial 0.033 0 0 change -x x x equilibrium 0.033 - x x x
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x = 5.77 ´ 10-4.
To find the pH,
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pH = -log [H3O+] = 10.76
The entire titration curve is shown below.
77. This is a disguised LeChâtelier's Principle
problem. The reaction is:
Cl2 (aq) + 2 H2O (l) « H3O+ (aq) + Cl- (aq) + HOCl (aq)
In order to minimize the amount of Cl-, the reaction equilibrium must be driven to the left - toward the reactants. To do this, one can increase the concentration of H3O+; the reaction will achieve equilibrium by partially consuming the additional hydronium. In so doing, some chloride and HOCl will also be consumed. Thus, the reaction should be run at low pH.
79. The reaction of urea, a base, with water will be:
H2O (l) + NH2CONH2 (aq) « OH- (aq) + NH2CONH3+ (aq)
In this reaction, it can be seen that the conjugate acid of urea is NH2CONH3+. The Ka of this species is:
This, by the way, is an exceptionally large Ka for a weak acid - it is almost a strong acid.A 0.15 M solution of this compound would result in the following equilibrium concentrations.
concentration NH2CONH3+ NH2CONH2 OH- initial 0.15 0 0 change -x x x equilibrium 0.15 - x x x
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x = 0.125 = [NH2CONH2]
(by the quadratic equation; you cannot make a simplifying assumption with such a large Ka)
Since x = [H3O+], pH = -log (0.125) = 0.90
80.
This is either a trick question or a poorly worded one. The question states that a solution of acetic acid and hydrochloric acid give the same color with methyl red indicator. Since methyl red is red at all pH values below ~4, we cannot be sure what the actual pH values for the solutions are.So, lets assume that both solutions have the same pH. In this case, the acetic acid will neutralize more sodium hydroxide than HCl. Consider the reaction,
The stoichiometry of the reaction is the same, regardless if HA is HCl or acetic acid. If the two solutions had the same pH, the concentration of acetic acid must be greater since it is a weak acid and HCl is a strong acid. In other words, it doesn't take as much of a strong acid to achieve the same hydronium ion concentration as a weak acid. Therefore, if the pH values are the same, the weak acid must have the greater concentration and, consequently, will neutralize a greater amount of base.
Of course the question does not explicitly state that the two solutions have the same pH. It could be the case that the acetic acid solution has a pH of 4 and the HCl could have a pH of -1. In this case, clearly, the HCl would neutralize more base.
85.
The total number of moles of the acid is 0.050; the total number of moles of the base is 0.060. Using the Henderson-Hasselbach equation, we have:
Sidebar: the above approach, while very good for most
weak acids and bases, has problems if Ka is too large. This
is because there is a substantial degree of ionization that will take place
even in the absence of added conjugate base - that is A- will
have two significant sources: the added A- and the ionization
of HA. A more rigorous approach is to set up a table of equilibrium concentrations
and solve as a normal equilibrium problem. That is,
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x = 0.0152 = [H3O+]
pH = -log (0.0152) = 1.82
This is significantly different than the result obtained from the Henderson-Hasselbach equation.
87.
All of the methods except "d" will make the desired buffer. That is because they all either start with a weak acid (acetic acid) and its conjugate base (acetate), or generate that combination via a partial neutralization reaction. Method d, although it results in a solution of the desired pH, is not a buffer because it does not contain a conjugate weak acid/base pair. It will therefore not be able to resist changes in pH in the same way as a buffer.
88.
This is a simple acid-base titration. You know volume and molarity of the added base. Therefore you know how many moles of OH- were added to reach the endpoint. You also know the reaction stoichiometry. Therefore you can write:
Additional Problems:
1a. This is a weak diprotic acid problem. The Ka values of carbonic acid are:
Ka1 = 4.3 ´ 10-71b. This is a diprotic base, with the following base ionization constants.Ka2 = 4.8 ´ 10-11
Since Ka1 >> Ka2, we can treat the solution in that same way as a monoprotic acid with Ka = Ka1. So, the reaction can be viewed simply as (H2A = H2CO3):
H2A + H2O « HA- + H3O+
We set up a table of equilibrium concentrations as follows:
concentration H2A HA- H3O+ initial 0.100 0 0 change -x x x equilibrium 0.100 - x x x
Since x = [H3O+], pH = -log (2.07 ´ 10-4) = 3.68x = 2.07 ´ 10-4
1c) NaHCO3 is an amphoteric compound. As such, a solution of NaHCO3 will have a pH given by:![]()
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Since Kb1 >> Kb2, this can be treated as a monoprotic base with Kb = Kb1. So, the reaction will be (B2- = CO32-)
B2- + H2O « BH- + OH-
concentration B2- BH- OH- initial 0.100 0 0 change -x x x equilibrium 0.100 - x x x
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x = [OH-] = 4.56 ´ 10-3
[H3O+] = 2.19 ´ 10-12; pH = 11.66
2.
To find the equilibrium concentrations of hydronium and hydroxide of very dilute solutions of acids and bases you need to consider the concentrations of these species that arise from the autoionization of water. Normally this source of hydronium and hydroxide is negligible compared to the products of acid or base ionization reactions.
Using the charge balance, we can say:a) for 1.0 ´ 10-6 M NaOH, we have:[H3O+] + [Na+] = [OH-]
The basis for this is the fact that the solution must remain neutral, that is, the sum of charge due to all of the cations must be balanced by the sum of all the charge due to the anions. The only ions present in the solution are Na+, H3O+ and OH-. Hydronium and hydroxide are related by Kw and we know the concentration of Na+ in each solution. We can then solve the equation for hydronium and determine the pH.
b) for 1.0 ´ 10-7 M NaOH, we have:![]()
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[H3O+]2 + (1 ´ 10-6)[H3O+] - Kw = 0
Using the quadratic equation yields, [H3O+] = 9.9 ´ 10-9; pH = 8.00
c) for 1.0 ´ 10-8 M NaOH, we have:![]()
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[H3O+]2 + (1 ´ 10-7)[H3O+] - Kw = 0
Using the quadratic equation yields, [H3O+] = 6.18 ´ 10-8; pH = 7.21
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[H3O+]2 + (1 ´ 10-8)[H3O+] - Kw = 0
Using the quadratic equation yields, [H3O+] = 9.51 ´ 10-8; pH = 7.02